Trig inverse derivatives, further understanding

[YehiaMedhat]

Given that: $\frac {d}{dx}f(x)= \frac {dy}{dx}$ and $\frac {d}{dx}f^-1(x)=\frac {dx}{dy}$, therefore, if $\frac {d}{dx}sin(x)=cos(x)$, therefore, $\frac {d}{dx}sin^-1(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}$, but, in my lectures it was: $\frac {d}{dx}sin^-1(x)=\frac {1}{\sqrt{1-x^2}}$
Is this similarity is my overthinking or there is a prove to how this $\frac {d}{dx}sin^-1(x)=\frac {1}{\sqrt{1-x^2}}$ came from $\frac {d}{dx}sin^-1(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}$
Please, if you have any refrences, mention it.

 

[pka]

I seems that you may be confusing notation.
The inverse trig functions are often written as an exponent.
The inverse sin is often given as $\sin^{-1}(x)\ne\dfrac{1}{\sin(x)}$
The correct notation is $\bf\arcsin(x)$ the derivative of which is [imath]\dfrac{1}{\sqrt{1-\sin^2(x)}[/imath]


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[Dr.Peterson]

Given that: $\frac {d}{dx}f(x)= \frac {dy}{dx}$ and $\frac {d}{dx}f^{-1}(x)=\frac {dx}{dy}$, therefore, if $\frac {d}{dx}sin(x)=cos(x)$, therefore, $\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}$, but, in my lectures it was: $\frac {d}{dx}sin^{-1}(x)=\frac {1}{\sqrt{1-x^2}}$
Is this similarity is my overthinking or there is a prove to how this $\frac {d}{dx}sin^{-1}(x)=\frac {1}{\sqrt{1-x^2}}$ came from $\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}$
Please, if you have any refrences, mention it.

I think you do understand that $\sin^{-1}(x)$ means the inverse, not the reciprocal; the reciprocals you are taking are [intended to be] of the derivative. But you're doing that incorrectly, by taking it too fast.

Your main error is in saying "$\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos(x)}$", as if the latter were $\frac {dx}{dy}$.

Writing it out more carefully, if $y=\sin^{-1}(x)$, then $x = \sin(y)$, so that $\frac{dx}{dy}=\cos(y)$.

Now we can properly take the reciprocal: $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\cos(y)}$$But this is in terms of y rather than x; so we need to express it in terms of x.

Since $\cos(y)=\sqrt{1-\sin^2(y)}=\sqrt{1-x^2}$, we can conclude that $$\frac {d}{dx}\sin^{-1}(x)=\frac{dy}{dx}=\frac{1}{\cos(y)}=\frac{1}{\sqrt{1-x^2}}$$
(We can take the positive sign because the inverse sine is an angle whose cosine is positive.)

 

[YehiaMedhat]

I think you do understand that $\sin^{-1}(x)$ means the inverse, not the reciprocal; the reciprocals you are taking are [intended to be] of the derivative. But you're doing that incorrectly, by taking it too fast.

Your main error is in saying "$\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos(x)}$", as if the latter were $\frac {dx}{dy}$.

Writing it out more carefully, if $y=\sin^{-1}(x)$, then $x = \sin(y)$, so that $\frac{dx}{dy}=\cos(y)$.

Now we can properly take the reciprocal: $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\cos(y)}$$But this is in terms of y rather than x; so we need to express it in terms of x.

Since $\cos(y)=\sqrt{1-\sin^2(y)}=\sqrt{1-x^2}$, we can conclude that $$\frac {d}{dx}\sin^{-1}(x)=\frac{dy}{dx}=\frac{1}{\cos(y)}=\frac{1}{\sqrt{1-x^2}}$$
(We can take the positive sign because the inverse sine is an angle whose cosine is positive.)

Okay, I seem to have mistakingly written the formulai
Thanks for explaination